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(N/A) Let us recall what a rectangle is.$A$ rectangle is a parallelogram in which one angle is a right angle.Let $ABCD$ be a rectangle in which $\angl

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(N/A) Let us recall what a rectangle is.
$A$ rectangle is a parallelogram in which one angle is a right angle.
Let $ABCD$ be a rectangle in which $\angle A = 90^{\circ}$.
We have to show that $\angle B = \angle C = \angle D = 90^{\circ}$.
We have $AD \parallel BC$ and $AB$ is a transversal.
So,$\angle A + \angle B = 180^{\circ}$ (Interior angles on the same side of the transversal are supplementary).
But,$\angle A = 90^{\circ}$.
So,$\angle B = 180^{\circ} - \angle A = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
Now,$\angle C = \angle A$ and $\angle D = \angle B$ (Opposite angles of a parallelogram are equal).
So,$\angle C = 90^{\circ}$ and $\angle D = 90^{\circ}$.
Therefore,each of the angles of a rectangle is a right angle.

Show that each angle of a rectangle is a right angle.

Similar Questions

$ABCD$ is a parallelogram in which $P$ and $Q$ are mid-points of opposite sides $AB$ and $CD$ (see Fig.). If $AQ$ intersects $DP$ at $S$ and $BQ$ intersects $CP$ at $R$,show that $PSQR$ is a parallelogram.

In parallelogram $ABCD$,two points $P$ and $Q$ are taken on diagonal $BD$ such that $DP = BQ$ (see Fig). Show that: $AP = CQ$.

$ABCD$ is a rhombus. Show that diagonal $AC$ bisects $\angle A$ as well as $\angle C$ and diagonal $BD$ bisects $\angle B$ as well as $\angle D$.

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In parallelogram $ABCD$,two points $P$ and $Q$ are taken on diagonal $BD$ such that $DP = BQ$ (see figure). Show that $APCQ$ is a parallelogram.

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